# Fundamental Theorem of Algebra (With Picard’s Little Theorem)

This post assumes familiarity with some basic concepts in complex analysis, including continuity and entire (everywhere complex-differentiable) functions. This is likely the simplest proof of the theorem (at least, among those that this author has seen), although it stands on the shoulders of a highly nontrivial theorem.

The fundamental theorem of algebra has quite a few number of proofs (enough to fill a book!). In fact, it seems a new tool in mathematics can prove its worth by being able to prove the fundamental theorem in a different way. This series of proofs of the fundamental theorem also highlights how in mathematics there are many many ways to prove a single theorem, and in re-proving an established theorem we introduce new concepts and strategies. And perhaps most of all, we accentuate the unifying beauty of mathematics.

Problem: Let $p(z): \mathbb{C} \to \mathbb{C}$ be a non-constant polynomial. Prove $p(z)$ has a root in $\mathbb{C}$.

Solution: Picard’s Little Theorem states that any entire function $\mathbb{C} \to \mathbb{C}$ whose range omits two distinct points must be constant. (See here for a proof sketch, and other notes)

Suppose to the contrary that $p$ has no zero. We claim $p$ also does not achieve some number in the set $\left \{ \frac{1}{k} : k \in \mathbb{N} \right \}$. Indeed, if it achieved all such values, we claim continuity would provide a zero. First, observe that as $p(z)$ is unbounded for large enough $|z|$, we may pick a $c \in \mathbb{R}$ such that $|p(z)| > 1$ whenever $|z| > c$. Then the points $z_k$ such that $p(z) = 1/k$ all lie within the closed disk of radius $c$. Because the set is closed and the complex plane is a complete metric space, we see that the sequence of $z_k$ converges to some $z$, and by continuity $p(z) = 0$.

In other words, no such sequence of $z_k$ can exist, and hence there is some $1/k$ omitted in the range of $p(z)$. As polynomials are entire, Picard’s Little theorem applies, and we conclude that $p(z)$ is constant, a contradiction. $\square$

## One thought on “Fundamental Theorem of Algebra (With Picard’s Little Theorem)”

1. Benjamin Bowman

“we see that the sequence of z_k converges to some z”. Isn’t it that z_k isn’t necessarily convergent but that it has a convergent subsequence by the Bolzanno Weirstrass theorem? Also, since we know no such sequence z_k can exist then there are only finitely many z_k and thus there are plenty of points omitted from the set of interest.